What is (x+1)^2+(y-3)^2=2 ?

The solution to (x+1)^2+(y-3)^2=2 is Circle with (a,b)=(-1,3),r=sqrt(2)

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(x+1)^2+(y-3)^2=2

(x + 1)^{2} + (y - 3)^{2} = 2

(a, b) = (- 1, 3), r = 2

Solution steps

(x + 1)^{2} + (y - 3)^{2} = 2

Rewrite

(x + 1)^{2} + (y - 3)^{2} = 2

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 3)^{2} = (2)^{2}

Therefore the circle properties are:

(a, b) = (- 1, 3), r = 2

\frac{( 2 x - 2 ) ^{2}}{2} + \frac{( 2 y + 6 ) ^{2}}{2} = \frac{46}{2}

x^{2} + y^{2} - 8 x + 10 y + 32 = 0

x^{2} + (y - \frac{16}{3})^{2} = 49

x^{2} + y^{2} - 2 x - 2 y + 4 + 5 - 64 = 0

x^{2} + 2 x + y^{2} - 5 = 0

What is (x+1)^2+(y-3)^2=2 ?
The solution to (x+1)^2+(y-3)^2=2 is Circle with (a,b)=(-1,3),r=sqrt(2)